2018 AMC 10A Problems/Problem 12
How many ordered pairs of real numbers satisfy the following system of equations?
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: Now, it becomes clear that there are intersection points. (pinetree1)
can be rewritten to . Substituting for in the second equation will give . Splitting this question into casework for the ranges of will give us the total number of solutions.
: will be negative so
is positive so and and
is negative so . and so there are no solutions ( can't equal to )
: It is fairly clear that
: will be positive so
will be negative so . There are no solutions (again, can't equal to )
will be positive so . and . Thus, the solutions are: , and the answer is . edit by pretzel, very minor edits by Bryanli, very very minor edit by ssb02
Note that can take on either of four values: , , , . Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: , , so the answer is
Just as in solution , we derive the equation . If we remove the absolute values, the equation collapses into four different possible values. , , , and , each equal to either or . Remember that if , then . Because we have already taken and into account, we can eliminate one of the conjugates of each pair, namely and , and and . Find the values of when , , and . We see that and give us the same value for , so the answer is
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